Software Engineer Interview Questions

Jennifer C. - "Reproduce the bug, read error messages, isolate the problem, use a debugger, check assumptions, review recent changes, consult documentation, Text Edge Cases, take break"See full answer

Nishigandha B. - "na"See full answer

Kwaku K. - "Use a BFS"See full answer

Lakshman B. - "As we can pass info to only one child at a time, I told that from any given node, we have to pass the info to that child(of this node) which has the largest subtree rooted at it. To calculate the subtree sizes, I used DFS. And then to calculate the minimum time to pass info to all the nodes, I used BFS picking the largest subtree child first at every node. I couldn't write the complete code in the given time and also made a mistake in telling the overall time complexity of my approach. I think t"See full answer

Idan R. - " read_dir(path: str) -> list[str] returns the full path of all files and sub- directories of a given directory. is_file(path: str) -> bool: returns true if the path points to a regular file. is_dir(path: str) -> bool: returns true if the path points to a directory. read_file(path: str) -> str: reads and returns the content of the file. The algorithm: notice that storing all the file contents' is too space intensive, so we can't read all the files' contents to store and compare with each"See full answer

Vince S. - "func isMatch(text: String, pattern: String) -> Bool { // Convert strings to arrays for easier indexing let s = Array(text.characters) let p = Array(pattern.characters) guard !s.isEmpty && !p.isEmpty else { return true } // Create DP table: dpi represents if s[0...i-1] matches p[0...j-1] var dp = Array(repeating: Array(repeating: false, count: p.count + 1), count: s.count + 1) // Empty pattern matches empty string dp[0]["See full answer

Srikar T. - "e"See full answer

Shar N. - "standard answer for this."See full answer

Ravi C. - "boolean isMatch(String s, String p) { int i=0; int j=0; int sID=-1; int prevM=-1; while(i<s.length()){ if(j<p.length() && (s.charAt(i)==p.charAt(j) || p.charAt(j)=='?')){ i++; j++; }else if(j<p.length() && p.charAt(j)=='*'){ sID=j; prevM=i; j++; }else if(sID!=-1){ j=sID+1; prevM++; i=prevM; }else{ return false; } } while(j<p.length() && p.charAt(j)=='*') j++; if(i!=s.length() || j!=p.leng"See full answer

Vishal T. - "Go has simpler syntax than Java. It is light weight. It is not Object Oriented. It does not support function overloading and function overriding. But these are small technical differences. Both are similar when it comes to testing. You have to create a mock object and implement an interface. Functionally, I did not feel any major difference."See full answer

Praveena P. - "I have been with Cisco, I am trying to move into a technical project manager role so that I will not loose the touch of latest technologies and also that aligns my passion."See full answer

Midde V. - "Front Page Layout Design for Newspaper App Header Section Logo: Displays at the top left. App Name: Displays alongside, so very prominent. Search Bar: Centered with search to find articles within the application. Navigation Menu: The links to the respective sections, World, Politics, Sports, Entertainment, and Opinion, etc Main Content Area Top Stories Carousel: It is a rotating banner that displays the top 3-5 news stories with images along with their headlines. Each story should be cl"See full answer

Aziz V. - "ArrayList allows constant time access (O(1)) to elements using their index because it uses a dynamic array internally, whereas LinkedList requires traversal from the head node, resulting in linear time complexity (O(n))."See full answer